A. Lisa is 15 years old and her father is 40. How many years ago was the father six times as old as Lisa?
B. Joel is one third the age of Bob. In 7 years the combined ages of the boys will be 58. How old is each boy now?
A. Let x = years ago
6(15-x) = 40-x
Solve for x.
B. B = 3J
(B+7) +( J+7) = 68
Substitute 3J for B in second equation and solve for J. Insert that value into the first equation and solve for B. Check by inserting both values into the second equation.
A. 10 yearsc
B. 11 and 33
A. Let's assume that the number of years ago is "x".
1. Lisa's current age is 15.
2. Her father's current age is 40.
If we subtract "x" years from both ages:
- Lisa's age would be 15 - x.
- Her father's age would be 40 - x.
According to the problem, her father was six times older than Lisa:
40 - x = 6(15-x)
To solve this equation, let's simplify it:
40 - x = 90 - 6x
Combine like terms:
5x = 50
Divide both sides of the equation by 5:
x = 10
Therefore, 10 years ago, Lisa's father was six times as old as Lisa.
B. Let's assume that Bob's current age is "x" and Joel's current age is "y".
According to the problem, Joel is one-third the age of Bob:
y = (1/3)x
In 7 years, the combined ages of the boys will be 58:
(x + 7) + (y + 7) = 58
Substitute the value of "y" from the first equation into the second equation:
(x + 7) + ((1/3)x + 7) = 58
Combine like terms:
x + (1/3)x + 14 = 58
To get rid of the fraction, multiply the entire equation by 3:
3x + x + 42 = 174
Combine like terms:
4x + 42 = 174
Subtract 42 from both sides of the equation:
4x = 132
Divide both sides of the equation by 4:
x = 33
Substitute the value of "x" back into the first equation:
y = (1/3) * 33
y = 11
Therefore, Bob is currently 33 years old, and Joel is currently 11 years old.
A. To solve this problem, we need to set up an equation. Let's assume that the number of years ago when the father was six times as old as Lisa is represented by x.
Currently, Lisa is 15 years old, so x years ago, she would have been 15 - x.
Similarly, the father is 40 years old now, so x years ago, he would have been 40 - x.
According to the problem, x years ago, the father was six times as old as Lisa. So we can set up the equation:
40 - x = 6(15 - x)
Now, we can solve this equation to find the value of x:
40 - x = 90 - 6x
Combine like terms:
5x = 50
Divide both sides by 5:
x = 10
Therefore, the father was six times as old as Lisa 10 years ago.
B. To solve this problem, let's assume that Bob's current age is represented by x. Since Joel is one-third the age of Bob, Joel's current age can be represented by (1/3)x.
In 7 years, Bob's age will be x + 7, and Joel's age will be (1/3)x + 7.
According to the problem, the combined ages of Bob and Joel in 7 years will be 58. So we can set up the equation:
(x + 7) + ((1/3)x + 7) = 58
Now, we can solve this equation to find the value of x:
(x + 1/3x) + 14 = 58
Combine like terms:
(4/3)x + 14 = 58
Subtract 14 from both sides:
(4/3)x = 44
Multiply both sides by 3/4:
x = 33
So Bob's current age is 33 years old, and Joel's current age is (1/3) * 33 = 11 years old.